# Centripital Force vs. Net Force; Inertia.

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#### Lunasly

##### Full Member
10+ Year Member
EDIT: I REVISED THIS POST. I KNOW ITS LONG BUT I'D REALLY APPRECIATE IT IF SOMEONE COULD PLEASE HELP ME CLEAR THIS UP!

I don't quite understand how we define centripetal force as I feel that the definition changes depending on whether we are talking about horizontal vs. vertical circular motion. If you don't mind, could you please critique my train of thought. I'm certain that I my explanation contains some inconsistencies, but I am not sure where my understanding is incorrect.

Here it goes:

It is to my understanding that vertical circular motion cannot be considered uniform given that the magnitude AND direction of the centripetal force acting on the object continually changes (and thus centripetal acceleration continually changes) depending on its position in its circular path. In other words, centripetal motion in the vertical direction is non-uniform.

I'd like to start off by defining what I think is centripetal force. Centripetal force (which I believe in not a real force) is the force placed on an object in order to maintain circular motion. The reason why I don't consider it a real force is because it is other forces (e.g., normal force, force of friction, tension, etc.) that constitute the centripetal force.

The particular question I having trouble with is a roller coaster cart that travels around a circular loop-de-loop (an example of vertical circular motion). I'd like to first define certain positions of the loop-de-loop to describe what I am talking about. Let the bottom of the loop be the 6 o'clock position, the top of the loop be the 12 o'clock position, and the right side of the loop (half way UP the loop) be the 3 o'clock position. At the top of the loop (12 o'clock position), the forces acting on the cart are the normal force (which points down) and the force of gravity (mg), which also points down. Many people claim that the normal force at this position is at its minimum (approaches 0) when we are considering the minimum speed required for the cart to complete the loop. In other words, it is the force of gravity that "provides" the centripetal force (i.e., the force required to maintain circular motion). Now, the reason I mention this is because I want to use this first example to define centripetal force. Both forces acting on the object are pointing straight down (which ALSO happens to be the centre of the circle) and so the NET force acting on the object is also down (and thus also towards the centre). Therefore, we can say that the NET force (which is the vector sum of the real component forces – i.e., force of gravity and normal force) is the centripetal force.

Now lets take a look at another instance in time. Lets imagine that the cart is at the 3 o'clock position (half way UP the loop). At this position, I do NOT see how we can equate the centripetal force with the net force acting on the object. The force which provides the centripetal force is the normal force (i.e., the normal force points directly inward towards the centre of the loop), however, the force of gravity now points downwards (as it always does) and not in the same direction as the normal force (as was the case when the cart was at the 12 o'clock position). Therefore, if the net force acting on the object is the vector sum of the component forces acting on the object (i.e., force of gravity and normal force), then the net force does not point towards the centre of the loop. In fact, the net force will point inwards, but it will not point towards the centre. If you need to see a picture of what this would look like, take a look at the figure "right side of loop" on this website (http://www.physicsclassroom.com/mmedia/circmot/rcd.cfm). My point is, we can still maintain circular motion because at any given position, the NET force always points INWARD towards the the centre of the circle. What I don't understand is: How is it that people always equate centripetal force (which always points towards the centre of the circle and is perpendicular to the tangential velocity of the object) to the net force acting on the object to solve centripetal motion problems when we can clearly see that at certain positions the centripetal force does not equal the net force acting on the object.

Other sources to consider: http://dev.physicslab.org/Document.aspx?doctype=3&filename=OscillatoryMotion_VerticalCircles.xml

I really appreciate your time in helping me figure this out. I am having a lot of trouble understanding this part of the dynamics unit.

Regards,
Lunasly.

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I just want to clarify a few concepts regarding centripetal motion. Thanks for the help! I am not talking about any specific passage, but this thought process of mine stems from the 2nd review passage of section 3 of TBR physics.

1) So after reading a few posts, apparently the net force is not equal to the centripetal force. The centripetal force is simply the sum of all of the forces that are required to maintain centripetal motion. For instance, at the 3 o' clock position of a roller coaster loop there are two forces:

a) The force of gravity (mg) points downwards (as always)
b) Centripetal force points inwards towards the centre of the loop.

With regard to (b), the collective forces (or just a single force) that is "acting" as the centripetal force is the normal force. In addition (and I might be wrong here), but the normal force is the equal and opposite force of the cart pushing down on the track. The reason why the cart pushes down on the track in the first place is because of the inertia, which as other might describe as the centrifugal force (maybe?).

Likewise, at the top of loop, since mg and normal force both point down, they are both the forces that make the centripetal force. This is not the case the 3 o'clock and 9 o' clock positions. Therefore, we can only set the sum of the forces to mv^2/r at the top of the loop and the bottom of the loop because it is only there that ALL of the forces constitute the centripetal force (I think?).

The net force is the resultant force of centripetal force (provided by the normal force) and the force of gravity. However, even though the net force is not equal to the centripetal force, the net force always points inwards. It has to point inwards in order to maintain circular motion (right)?

2) Another question I have is with regard to the speed of a cart at certain points of the roller coaster. Why is the speed of a roller coaster cart highest at the 3 o' clock and 9 o' clock positions? I read this in another post and could not understand the reasoning behind it.
For circular motion, the net force is actually the centripetal force. This is the key to solving any circular motion problems, including the roller coaster example you gave. For instance, at the bottom of the rollercoaster, you know you have the weight of the coaster pointing down and the normal force pointing up. The sum of these forces equals the centripetal force. Because the net force (centripetal force) points towards the center, and because at this point the normal force is also pointing up and towards the center, then by convention they must share the same sign. So mathematically you'd describe this as:

Fnet = Fnormal - Fweight ; What's important is you choose the same convention for directionality. I chose Fnet and Fnormal to be positive for simplicity. And because Fnet = mv^2/r (centripetal force), then we can say: mv^2/r = Fnormal - Fweight. If a question asked you what the apparent weight is at the bottom of the roller coaster (Ie. would you feel heavier or lighter), then you can re-arrange to figure it out. Whenever you're asked to find the apparent weight, it's the same thing as normal force.

Therefore:

Fnet + Fweight = Fnormal (or mv^2/r + mg = Fnormal). So as you might expect, we'd feel heavier at the bottom (by an amount equal to mv^2/r).

Right, that makes sense, but the centripetal force is still the net force at the 3 o clock position? Other threads I have read would disagree. And does centripetal force always = net force when are talking about horizontal vs. Vertical centripetal motion?

Right, that makes sense, but the centripetal force is still the net force at the 3 o clock position? Other threads I have read would disagree. And does centripetal force always = net force when are talking about horizontal vs. Vertical centripetal motion?
The centripetal force is always due to some other naturally occuring force. At the 3' oclock position, the only force pointing directly towards the center would be the normal force due the roller coaster. Therefore, since this is technically the only force at that position, it equals the centripetal force. However, there is also a net force in the y-direction due to the downward weight of the coaster. This force acts to accelerate the coaster at that position, but does not effect the net centripetal force pointing inwards.

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